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2073. Time Needed to Buy Tickets

Description

There are n people in a line queuing to buy tickets, where the 0th person is at the front of the line and the (n - 1)th person is at the back of the line.

You are given a 0-indexed integer array tickets of length n where the number of tickets that the ith person would like to buy is tickets[i].

Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.

Return the time taken for the person at position k (0-indexed) to finish buying tickets.

 

Example 1:

Input: tickets = [2,3,2], k = 2
Output: 6
Explanation: 
- In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].
- In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].
The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.

Example 2:

Input: tickets = [5,1,1,1], k = 0
Output: 8
Explanation:
- In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0].
- In the next 4 passes, only the person in position 0 is buying tickets.
The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.

 

Constraints:

  • n == tickets.length
  • 1 <= n <= 100
  • 1 <= tickets[i] <= 100
  • 0 <= k < n

Solutions

  • class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int ans = 0;
        for (int i = 0; i < tickets.length; i++) {
            if (i <= k) {
                ans += Math.min(tickets[k], tickets[i]);
            } else {
                ans += Math.min(tickets[k] - 1, tickets[i]);
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int timeRequiredToBuy(vector<int>& tickets, int k) {
        int ans = 0;
        for (int i = 0; i < tickets.size(); ++i) {
            if (i <= k) {
                ans += min(tickets[k], tickets[i]);
            } else {
                ans += min(tickets[k] - 1, tickets[i]);
            }
        }
        return ans;
    }
};

  • class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        ans = 0
        for i, t in enumerate(tickets):
            if i <= k:
                ans += min(tickets[k], t)
            else:
                ans += min(tickets[k] - 1, t)
        return ans


  • func timeRequiredToBuy(tickets []int, k int) int {
	ans := 0
	for i, t := range tickets {
		if i <= k {
			ans += min(tickets[k], t)
		} else {
			ans += min(tickets[k]-1, t)
		}
	}
	return ans
}

  • function timeRequiredToBuy(tickets: number[], k: number): number {
    const n = tickets.length;
    let target = tickets[k] - 1;
    let ans = 0;
    // round1
    for (let i = 0; i < n; i++) {
        let num = tickets[i];
        if (num <= target) {
            ans += num;
            tickets[i] = 0;
        } else {
            ans += target;
            tickets[i] -= target;
        }
    }

    // round2
    for (let i = 0; i <= k; i++) {
        let num = tickets[i];
        ans += num > 0 ? 1 : 0;
    }
    return ans;
}

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